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_bTZbn7M2Hc
|
So this becomes the sum
over k different from n,
|
_bTZbn7M2Hc
|
because for k equal to n, these
are orthogonal of k0 k0 n1.
|
_bTZbn7M2Hc
|
And that's what we
calculated here.
|
_bTZbn7M2Hc
|
So what did we get?
|
_bTZbn7M2Hc
|
Therefore, the state n1, I can
substitute what we had there.
|
_bTZbn7M2Hc
|
It's the sum from k different
from n of k0 delta Hnk over Ek0
|
_bTZbn7M2Hc
|
minus En0.
|
_bTZbn7M2Hc
|
That's n1.
|
_bTZbn7M2Hc
|
I should have a minus sign.
|
_bTZbn7M2Hc
|
The minus sign is
there at the state n1.
|
_bTZbn7M2Hc
|
So the state n1 is a
complicated correction.
|
_bTZbn7M2Hc
|
It gets a little component
from every other state
|
_bTZbn7M2Hc
|
of the spectrum.
|
_bTZbn7M2Hc
|
And the coefficient depends
on the matrix element
|
_bTZbn7M2Hc
|
of your state with the state
you're contributing with.
|
_bTZbn7M2Hc
|
So you have the state n and
all the other states here.
|
_bTZbn7M2Hc
|
The amount of this state k
that enters into the correction
|
_bTZbn7M2Hc
|
is proportional to the matrix
element between n and k.
|
_bTZbn7M2Hc
|
If the matrix element is 0, that
state does not contribute here.
|
_bTZbn7M2Hc
|
And then there is the
energy denominator as well.
|
_bTZbn7M2Hc
|
So we're getting to the
end of this calculation.
|
_bTZbn7M2Hc
|
There's one more thing one
can do, which is to find--
|
_bTZbn7M2Hc
|
so I'm starting to
wrap up this, but still
|
_bTZbn7M2Hc
|
an important step
what we have to do.
|
_bTZbn7M2Hc
|
I'll get the second
order energy correction.
|
_bTZbn7M2Hc
|
What is our second
order energy correction?
|
_bTZbn7M2Hc
|
Our second order
energy correction
|
_bTZbn7M2Hc
|
can be found from the formula
on that blackboard, En2.
|
_bTZbn7M2Hc
|
We already found the first
order energy correction,
|
_bTZbn7M2Hc
|
which I happened to have
erased it right now.
|
_bTZbn7M2Hc
|
It was there.
|
_bTZbn7M2Hc
|
En2 is obtained by doing
n0 delta H times n1, which
|
_bTZbn7M2Hc
|
we already know.
|
_bTZbn7M2Hc
|
So I must do n0 delta H on that.
|
_bTZbn7M2Hc
|
So look what you get.
|
_bTZbn7M2Hc
|
You get minus the sum
over k different from n.
|
_bTZbn7M2Hc
|
Think of putting the n0 and the
delta H, they're all together.
|
_bTZbn7M2Hc
|
It's a [INAUDIBLE] so far.
|
_bTZbn7M2Hc
|
It's a delta H and n0.
|
_bTZbn7M2Hc
|
IT should go into n1.
|
_bTZbn7M2Hc
|
But the only state in n1 is k0.
|
_bTZbn7M2Hc
|
So here we have k0.
|
_bTZbn7M2Hc
|
And then we have delta
Hnk over Ek0 minus En0.
|
_bTZbn7M2Hc
|
OK.
|
_bTZbn7M2Hc
|
A little bit of work.
|
_bTZbn7M2Hc
|
So what is this?
|
_bTZbn7M2Hc
|
This is another matrix element.
|
_bTZbn7M2Hc
|
This is the matrix--
|
_bTZbn7M2Hc
|
OK.
|
_bTZbn7M2Hc
|
I'm sorry.
|
_bTZbn7M2Hc
|
Here do I have a mistake?
|
_bTZbn7M2Hc
|
Oh, yes, I have kn.
|
_bTZbn7M2Hc
|
I copied it wrong.
|
_bTZbn7M2Hc
|
It's kn.
|
_bTZbn7M2Hc
|
Yes.
|
_bTZbn7M2Hc
|
Yes.
|
_bTZbn7M2Hc
|
So here I have delta Hnk.
|
_bTZbn7M2Hc
|
but delta Hnk is this.
|
_bTZbn7M2Hc
|
If you complex conjugate--
|
_bTZbn7M2Hc
|
if you complex
conjugate delta Hkn,
|
_bTZbn7M2Hc
|
complex conjugate is k
delta H n complex conjugate,
|
_bTZbn7M2Hc
|
which changes the order.
|
_bTZbn7M2Hc
|
n delta H, which
is her mission k.
|
_bTZbn7M2Hc
|
And that's delta Hnk.
|
_bTZbn7M2Hc
|
So delta Hnk is equal
to delta Hkn star.
|
_bTZbn7M2Hc
|
And therefore the second
order energy correction
|
_bTZbn7M2Hc
|
has a nice formula.
|
_bTZbn7M2Hc
|
En2 is equal to minus the
sum over k different from n.
|
_bTZbn7M2Hc
|
Delta Hnk, which is the
star of that times this one,
|
_bTZbn7M2Hc
|
so you get delta Hnk absolute
value squared divided by Ekn.
|
_bTZbn7M2Hc
|
Ek0 minus En0.
|
_bTZbn7M2Hc
|
So we've done a lot of work.
|
_bTZbn7M2Hc
|
We've written the perturbation.
|
_bTZbn7M2Hc
|
Here is the answer.
|
_bTZbn7M2Hc
|
So far we have n of lambda
equal n0 plus lambda n1.
|
_bTZbn7M2Hc
|
n1 has been calculated.
|
_bTZbn7M2Hc
|
Energy is En0 plus lambda En1.
|
_bTZbn7M2Hc
|
That was calculated
what was just
|
_bTZbn7M2Hc
|
delta H in this state
plus lambda squared
|
_bTZbn7M2Hc
|
En2, which we have calculated.
|
_bTZbn7M2Hc
|
So this is as far as we will do
for nondegenerate perturbation
|
_bTZbn7M2Hc
|
theory.
|
_bTZbn7M2Hc
|
But we have found rather
interesting formulas.
|
_bTZbn7M2Hc
|
And we're going to spend
half of its lecture trying
|
_bTZbn7M2Hc
|
to understand them better.
|
rLlZpnT02ZU
|
The following content is
provided under a Creative
|
rLlZpnT02ZU
|
Commons license.
|
rLlZpnT02ZU
|
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|
rLlZpnT02ZU
|
continue to offer high quality
educational resources for free.
|
rLlZpnT02ZU
|
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view additional materials
|
rLlZpnT02ZU
|
from hundreds of MIT courses,
visit MIT OpenCourseWare
|
rLlZpnT02ZU
|
at ocw.mit.edu.
|
rLlZpnT02ZU
|
PHILIPPE RIGOLLET: --124.
|
rLlZpnT02ZU
|
If I were to repeat
this 1,000 times,
|
rLlZpnT02ZU
|
so every one of
those 1,000 times
|
rLlZpnT02ZU
|
they collect 124
data points and then
|
rLlZpnT02ZU
|
I'd do it again and
do it again and again,
|
rLlZpnT02ZU
|
then in average, the
number I should get
|
rLlZpnT02ZU
|
should be close to the true
parameter that I'm looking for.
|
rLlZpnT02ZU
|
The fluctuations that
are due to the fact
|
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